Question

find derivative of (cscx)/(3sinx)
I got (3cot+3)/(3sinx)^2 but I need verification.

Answer 1

\[(-2/3) \cot(x) \csc^2(x)\] is what I get.

Factor out constants:

1/3 (d/dx(csc^2(x)))

Use the chain rule, d/dx(csc^2(x)) = ( du^2)/( du) ( du)/( dx), where u = csc(x) and ( du^2)/( du) = 2 u:

1/3 (2 csc(x) (d/dx(csc(x))))

The derivative of csc(x) is -cot(x) csc(x):

-2/3 csc(x) (-cot(x) csc(x))

Answer 2

we haven't learned chain rule yet, is there ayway to do it without chain rule?

Answer 3

Have you learnt the quotient rule?

Answer 4

yes

Answer 5

sin x cosec x =1
cosec x cos x = cot x
Did you use these?

Answer 6

Try taking a factor of (1/3) out of the equation before you use the quotient rule also, this may help.

Answer 7

i dotn understand how you got that. this is what I did, I took out the constant 1/3, so now I have cscx/sin....so (sin)(-cscxcotx)-(cscx)(cosx)/(sinx)^2

Answer 8

yep, you can simplify what you have, remember cosec x is just 1/sin x, so it will simplify to -cot x on the LHS of the subtraction sign...

Answer 9

yeah I understand that part, now I have (-cotx-cscxcosx)/(sin)^2....so what else would i simplify

Answer 10

since cosec is 1/sin, you have cos/sin which is cot, now you have cot cot which is cot^2

Answer 11

the original answer is just rearranged slightly, you're good

Answer 12

do you mean -cot^2?